Electrochemistry

Standard hydrogen electrode (SHE)

The SHE is the primary reference electrode. By convention its electrode potential is taken as exactly 0.00 V at all temperatures, and the potentials of all other electrodes are measured against it.

Construction

1. A small piece of platinum foil coated with platinum black (finely divided Pt) is fused to a platinum wire sealed in a glass tube. The wire makes contact with the outer circuit, often through mercury in the tube. Platinum black adsorbs hydrogen and speeds up the equilibrium between $\mathrm{H_2}$ and $\mathrm{H^+}$.
2. The foil is surrounded by an outer glass jacket with an inlet for hydrogen gas near the top and openings at the bottom.
3. The whole assembly is dipped in a solution with unit concentration of $\mathrm{H^+}$ ions (1 M HCl, activity = 1).
4. Pure, dry hydrogen gas at 1 atm pressure is bubbled continuously over the foil at 298 K (25 °C).

                 H2 gas (pure, 1 atm)
                        |
                        v
          Pt wire --> +-|-+----------+
          (to circuit)| | |          |  <- outer glass jacket
                      | | |   H2     |
     -----------------|-|-|----------|--------
       1 M HCl        | | |          |   (a H+ = 1, 298 K)
                      | |_|          |
                      | [###]  Pt foil coated
                      | [###]  with Pt black
                      |  o  o  o  o  |  <- H2 bubbles escape
                      +--------------+

Representation: $\mathrm{Pt,\ H_2(g,\ 1\ atm)\ |\ H^+(1\ M)}$, with $E^\circ = 0.00$ V.

Working

The electrode sets up an equilibrium between hydrogen gas and hydrogen ions on the platinum surface:

$$\mathrm{2H^+(aq) + 2e^- \rightleftharpoons H_2(g)}, \qquad E^\circ = 0.00\ \text{V}$$

The SHE is reversible, so it can act as either anode or cathode, depending on the electrode it is joined to.

• As anode (when joined to an electrode with a positive potential, e.g. Cu): oxidation takes place.

$$\mathrm{H_2(g) \longrightarrow 2H^+(aq) + 2e^-}$$

• As cathode (when joined to an electrode with a negative potential, e.g. Zn): reduction takes place.

$$\mathrm{2H^+(aq) + 2e^- \longrightarrow H_2(g)}$$

Measurement of an electrode potential. The test electrode is joined to the SHE through a salt bridge and the emf is measured. Because $E^\circ_{SHE} = 0$, the measured emf gives the electrode potential directly.

Example: $\mathrm{Zn\,|\,Zn^{2+}(1\,M)\,\|\,H^+(1\,M)\,|\,H_2(1\,atm),\,Pt}$ gives emf = 0.76 V, with zinc as the anode:

$$E^\circ_{cell} = E^\circ_{SHE} - E^\circ_{Zn^{2+}/Zn} \Rightarrow 0.76 = 0 - E^\circ_{Zn^{2+}/Zn} \Rightarrow E^\circ_{Zn^{2+}/Zn} = -0.76\ \text{V}$$

Similarly, Cu with the SHE gives emf 0.34 V with Cu as the cathode, so $E^\circ_{Cu^{2+}/Cu} = +0.34$ V.

Limitations of the hydrogen electrode

1. It is hard to keep the pressure of $\mathrm{H_2}$ at exactly 1 atm and the $\mathrm{H^+}$ concentration (activity) at exactly 1 M throughout.
2. Pure, dry hydrogen gas is difficult to prepare and maintain.
3. The platinum black is easily poisoned by impurities in the gas or solution (e.g. arsenic, sulphur compounds and other adsorbed substances), which makes it inactive.
4. It cannot be used in solutions containing oxidising or reducing agents, or ions of more easily reduced metals, which react with $\mathrm{H_2}$ or with the platinum.
5. It is bulky, not portable, and platinum is expensive.

Because of these difficulties, secondary reference electrodes such as the calomel electrode are generally used in practice.

i) Electrochemical cell

An electrochemical (galvanic or voltaic) cell is a device that turns the chemical energy of a spontaneous redox reaction into electrical energy. Oxidation takes place at the anode and reduction at the cathode, in two separate half-cells joined by a salt bridge and an external wire. Example: the Daniell cell.

ii) Representation of the cell

Al has the lower reduction potential (−1.67 V), so it is the anode. Ag (+0.80 V) is the cathode.

$$\underset{\text{Anode }(-)}{\mathrm{Al(s)\,|\,Al^{3+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{Ag^{+}(1\,M)\,|\,Ag(s)}}$$

iii) Complete cell reactions

At anode (oxidation):

$$\mathrm{Al(s) \longrightarrow Al^{3+}(aq) + 3e^-}$$

At cathode (reduction), multiplied by 3 to balance the electrons:

$$\mathrm{3Ag^{+}(aq) + 3e^- \longrightarrow 3Ag(s)}$$

Net cell reaction:

$$\mathrm{Al(s) + 3Ag^{+}(aq) \longrightarrow Al^{3+}(aq) + 3Ag(s)}$$

iv) EMF of the cell

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-1.67) = +2.47\ \text{V}$$

The positive emf shows that the cell reaction is spontaneous.

a) Construction of the galvanic cell

Iron has the lower reduction potential (−0.44 V), so it is oxidised and is the anode. Copper (+0.34 V) is reduced and is the cathode.

• An iron rod is dipped in 1 M $\mathrm{FeSO_4}$ solution (anode half-cell).
• A copper rod is dipped in 1 M $\mathrm{CuSO_4}$ solution (cathode half-cell).
• The two solutions are joined by a salt bridge (KCl or $\mathrm{KNO_3}$ in agar), and the rods are joined through a voltmeter. Electrons flow from Fe to Cu in the outer circuit.

              e-  ------------>
       +-------------( V )-------------+
       |                               |
   Fe rod (anode, -)            Cu rod (cathode, +)
       |      ________________         |
    ___|___  |  salt bridge   |  ______|__
   |   |   | |                | |   |     |
   |   |   |_|                |_|   |     |
   |  [Fe] |                    |  [Cu]   |
   |_______|                    |_________|
   1 M FeSO4 (Fe2+)            1 M CuSO4 (Cu2+)

• Anode: $\mathrm{Fe \longrightarrow Fe^{2+} + 2e^-}$
• Cathode: $\mathrm{Cu^{2+} + 2e^- \longrightarrow Cu}$
• Cell reaction: $\mathrm{Fe + Cu^{2+} \longrightarrow Fe^{2+} + Cu}$

Cell notation:

$$\underset{\text{Anode }(-)}{\mathrm{Fe(s)\,|\,Fe^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{Cu^{2+}(1\,M)\,|\,Cu(s)}}$$

b) Standard emf

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.44) = +0.78\ \text{V}$$

c) Why the blue colour of $\mathrm{CuSO_4}$ disappears

• The blue colour of $\mathrm{CuSO_4}$ solution comes from the hydrated $\mathrm{Cu^{2+}}$ ion, $\mathrm{[Cu(H_2O)_6]^{2+}}$.
• Iron has a lower reduction potential (−0.44 V) than copper (+0.34 V), so iron is the stronger reducing agent. It sits above copper in the electrochemical series and displaces copper from its salt solution:

$$\mathrm{Fe(s) + CuSO_4(aq) \longrightarrow FeSO_4(aq) + Cu(s)}$$

• This reaction is spontaneous: $E^\circ = +0.78$ V and $\Delta G^\circ = -nFE^\circ < 0$.
• As the $\mathrm{Cu^{2+}}$ ions are reduced to metallic copper, which deposits as a reddish-brown coat on the iron rod, the blue colour fades. The solution becomes pale green because of the $\mathrm{Fe^{2+}}$ ions formed.

(a) Li-ion battery

Laptops and mobile phones use rechargeable lithium-ion batteries. Lithium is very light and has a very negative electrode potential, so these batteries give a high voltage (about 3.7 V) and a high energy density in a small, light pack, and they can be recharged many times.

Zinc has a negative reduction potential (−0.76 V), lower than that of SHE (0.00 V). So zinc is the anode (negative) and SHE is the cathode (positive).

i) Diagram of the galvanic cell

                 e-  -------------->
        +---------------( V )------------------+
        |                                      |
   Zn rod (anode, -)               Pt (cathode, +)   H2 gas in
        |                                      |      | (1 atm)
        |        salt bridge (KCl)             |      v
     ___|___   _____________________   ________|______|__
    |   |   | |                     | |  |     |      |  |
    |   |   |_|                     |_|  |     |      |  |
    |   |   |                         |  |   [Pt]     |  |
    |  [Zn] |                         |  | platinised |  |
    |       |                         |  |  o  o  o  o   |  <- H2 bubbles
    |_______|                         |__________________|
   1 M ZnSO4 (Zn2+)                     1 M HCl (H+), 298 K

Cell notation:

$$\underset{\text{Anode }(-)}{\mathrm{Zn(s)\,|\,Zn^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{H^{+}(1\,M)\,|\,H_2(g,\,1\,atm)\,|\,Pt}}$$

ii) Complete cell reaction

At anode (oxidation):

$$\mathrm{Zn(s) \longrightarrow Zn^{2+}(aq) + 2e^-}$$

At cathode (reduction):

$$\mathrm{2H^{+}(aq) + 2e^- \longrightarrow H_2(g)}$$

Net cell reaction:

$$\mathrm{Zn(s) + 2H^{+}(aq) \longrightarrow Zn^{2+}(aq) + H_2(g)}$$

iii) EMF of the cell

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{SHE} - E^\circ_{Zn^{2+}/Zn} = 0.00 - (-0.76) = +0.76\ \text{V}$$

(d) S > Q > R > P

The lower (more negative) the standard reduction potential, the more easily the species loses electrons, so the stronger it is as a reducing agent. In order of reduction potential: S (−3.05 V) < Q (−0.76 V) < R (0.00 V) < P (+0.34 V). Reducing power therefore runs S > Q > R > P.

(a) Standard electrode potential

The standard electrode potential ($E^\circ$) is the potential of an electrode measured against the standard hydrogen electrode (taken as 0.00 V) under standard conditions: 1 M ion concentration, 1 atm pressure for gases and 25 °C (298 K). By IUPAC convention it is written as a reduction potential.

(b) Cell notation

Mg has the lower reduction potential (−2.38 V), so it is oxidised and is the anode. Zn (−0.76 V) is the cathode.

$$\underset{\text{Anode }(-)}{\mathrm{Mg(s)\,|\,Mg^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{Zn^{2+}(1\,M)\,|\,Zn(s)}}$$

(c) Complete cell reaction

• Anode: $\mathrm{Mg(s) \longrightarrow Mg^{2+}(aq) + 2e^-}$
• Cathode: $\mathrm{Zn^{2+}(aq) + 2e^- \longrightarrow Zn(s)}$

$$\mathrm{Mg(s) + Zn^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Zn(s)}$$

(d) Cell potential at 25 °C

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.76 - (-2.38) = +1.62\ \text{V}$$

Copper has the lower reduction potential (+0.34 V), so it is the anode (oxidation, negative). Silver (+0.80 V) is the cathode (reduction, positive).

(a) Diagram of the electrochemical cell

                 e-  -------------->
        +---------------( V )----------------+
        |                                    |
   Cu rod (anode, -)                  Ag rod (cathode, +)
        |       salt bridge (KNO3)           |
     ___|___   ____________________   _______|__
    |   |   | |                    | |   |      |
    |   |   |_|                    |_|   |      |
    |   |   |   <-- NO3-   K+ -->    |   |      |
    |  [Cu] |                        |  [Ag]    |
    |_______|                        |__________|
   1 M CuSO4 (Cu2+)                 1 M AgNO3 (Ag+)

Cell notation:

$$\underset{\text{Anode }(-)}{\mathrm{Cu(s)\,|\,Cu^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{Ag^{+}(1\,M)\,|\,Ag(s)}}$$

(b) Function of the salt bridge

• It completes the inner circuit by letting ions move between the two half-cells, without the two solutions mixing.
• It keeps both solutions electrically neutral. $\mathrm{NO_3^-}$ ions move into the anode compartment to balance the extra $\mathrm{Cu^{2+}}$ formed there, and $\mathrm{K^+}$ ions move into the cathode compartment to replace the $\mathrm{Ag^+}$ used up. Without it the current would stop almost at once.
• It reduces the liquid junction potential. ($\mathrm{KNO_3}$ is used here instead of KCl because $\mathrm{Cl^-}$ would precipitate AgCl.)

(c) Cell reaction and emf

• Anode: $\mathrm{Cu(s) \longrightarrow Cu^{2+}(aq) + 2e^-}$
• Cathode: $\mathrm{2Ag^{+}(aq) + 2e^- \longrightarrow 2Ag(s)}$

Net cell reaction:

$$\mathrm{Cu(s) + 2Ag^{+}(aq) \longrightarrow Cu^{2+}(aq) + 2Ag(s)}$$

EMF:

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - 0.34 = +0.46\ \text{V}$$

(c) $\mathrm{Zn + Cd^{2+} \longrightarrow Zn^{2+} + Cd}$

In the notation the left electrode (Zn) is the anode, where oxidation takes place: $\mathrm{Zn \to Zn^{2+} + 2e^-}$. The right electrode (Cd) is the cathode, where reduction takes place: $\mathrm{Cd^{2+} + 2e^- \to Cd}$. Adding the two gives the cell reaction. It is spontaneous because $E^\circ_{cell} = -0.40 - (-0.76) = +0.36$ V.

(a) Power for heat and light

In spacecraft such as Apollo, the $\mathrm{H_2}$–$\mathrm{O_2}$ fuel cell produced electricity for the on-board systems (lighting, heating, instruments and communication). Its product, pure water, was used as drinking water for the astronauts:

$$\mathrm{2H_2(g) + O_2(g) \longrightarrow 2H_2O(l)}$$

It is not used for propulsion or lift; those come from rocket engines.

Why SHE is called a reference electrode

The absolute potential of a single electrode cannot be measured. So the potential of the standard hydrogen electrode is arbitrarily fixed as 0.00 V at all temperatures, and the potentials of all other electrodes are measured relative to it. Because it is the standard against which other electrode potentials are measured, it is called a reference electrode.

i) Cell notation

The emf is +0.76 V and Zn has a negative potential relative to SHE, so zinc is oxidised. Zn is the anode (negative) and SHE is the cathode (positive).

$$\underset{\text{Anode }(-)}{\mathrm{Zn(s)\,|\,Zn^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{H^{+}(1\,M)\,|\,H_2(g,\,1\,atm)\,|\,Pt}}$$

ii) Net cell reaction

• Anode: $\mathrm{Zn \longrightarrow Zn^{2+} + 2e^-}$
• Cathode: $\mathrm{2H^+ + 2e^- \longrightarrow H_2}$

$$\mathrm{Zn(s) + 2H^{+}(aq) \longrightarrow Zn^{2+}(aq) + H_2(g)}$$

iii) Value of $E^\circ_{Zn^{2+}/Zn}$

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$ $$+0.76 = 0.00 - E^\circ_{Zn^{2+}/Zn}$$ $$E^\circ_{Zn^{2+}/Zn} = -0.76\ \text{V}$$

Mg has the lower (more negative) reduction potential, so it is oxidised and forms the anode. Cu has the higher reduction potential, so $Cu^{2+}$ is reduced at the cathode.

a) Cell notation [1]

$$Mg(s)\,|\,Mg^{2+}(1\,M)\,\|\,Cu^{2+}(1\,M)\,|\,Cu(s)$$

Anode (left): Mg. Cathode (right): Cu.

b) Standard emf [2]

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (+0.34) - (-2.37) = +2.71\ V$$

c) Cell reaction [2]

Anode (oxidation): $Mg \rightarrow Mg^{2+} + 2e^-$

Cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu$

$$Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$$

The positive emf shows the reaction is spontaneous.

(c) +2.71 V

Mg is oxidised (anode) and $Cu^{2+}$ is reduced (cathode), so

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-2.37) = +2.71\ \text{V}$$

Calomel electrode. It is a metal-metal salt electrode consisting of mercury in contact with mercurous chloride (calomel, $Hg_2Cl_2$) and a KCl solution, with a platinum wire for contact. Notation: $Hg|Hg_2Cl_2|Cl^-(aq)$. Half-reaction:

$$Hg_2Cl_2 + 2e^- \rightarrow 2Hg + 2Cl^-$$

Why a secondary reference electrode? The primary reference electrode is the standard hydrogen electrode (SHE), whose potential is defined as 0 V. SHE is awkward to set up (pure $H_2$ at 1 atm, platinised Pt, 1 M $H^+$). The calomel electrode is easy to prepare and handle, and its potential (about +0.24 V for saturated KCl) was fixed by comparison with SHE. Since its potential is measured against SHE, it is a secondary reference electrode.

(a) $E^\circ(Fe^{2+}/Fe) = -0.44$ V is lower, so Fe is the anode (oxidation); $E^\circ(Cu^{2+}/Cu) = +0.34$ V is higher, so Cu is the cathode (reduction).

• Anode: $Fe \rightarrow Fe^{2+} + 2e^-$
• Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$

Cell notation:

$$Fe(s)\,|\,Fe^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)$$

(b)

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.44) = +0.78\ \text{V}$$

(d) $\mathrm{H_2SO_4}$ is consumed

During discharge both electrodes change to lead sulphate and water is formed:

$$\mathrm{Pb + PbO_2 + 2H_2SO_4 \longrightarrow 2PbSO_4 + 2H_2O}$$

So sulphuric acid is used up and the density of the electrolyte falls. $\mathrm{PbSO_4}$ is formed, not consumed, and no $\mathrm{SO_2}$ is given off. Lead is used up at the anode, not formed.

a. Cell notation

Copper has the lower reduction potential (+0.34 V), so it is oxidised and acts as the anode (negative). Silver (+0.80 V) is reduced and acts as the cathode (positive).

$$\underset{\text{Anode }(-)}{\mathrm{Cu(s)\,|\,Cu^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{Ag^{+}(1\,M)\,|\,Ag(s)}}$$

• Anode: $\mathrm{Cu \longrightarrow Cu^{2+} + 2e^-}$
• Cathode: $\mathrm{2Ag^+ + 2e^- \longrightarrow 2Ag}$
• Cell reaction: $\mathrm{Cu + 2Ag^+ \longrightarrow Cu^{2+} + 2Ag}$

b. Cell potential (standard conditions: 1 M, 25 °C, 1 atm)

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu} = 0.80 - 0.34 = +0.46\ \text{V}$$

c. Free energy change

Given: $n = 2$, $F = 96500\ \text{C mol}^{-1}$, $E^\circ_{cell} = 0.46\ \text{V}$

$$\Delta G^\circ = -nFE^\circ_{cell} = -2 \times 96500 \times 0.46 = -88780\ \text{J mol}^{-1} \approx -88.78\ \text{kJ mol}^{-1}$$

Since $\Delta G^\circ$ is negative, the reaction is spontaneous.

d. Can $\mathrm{AgNO_3}$ be stored in a copper vessel?

No. Copper has a lower reduction potential than silver, so copper is the stronger reducing agent. It would react with the $\mathrm{Ag^+}$ ions:

$$\mathrm{Cu(s) + 2AgNO_3(aq) \longrightarrow Cu(NO_3)_2(aq) + 2Ag(s)}, \quad E^\circ = +0.46\ \text{V}\ (\Delta G^\circ < 0)$$

The vessel would dissolve (corrode), silver would deposit on it, and the solution would turn blue.

i) Activity series

• B displaces only A from solution, so B is more reactive than A, and B is less reactive than C and D.
• Only D and C displace hydrogen from 1 M HCl, so C and D lie above hydrogen, while A and B lie below it.
• No metal displaces C, so C is the most reactive.

$$\mathrm{C > D > H > B > A}$$

Reactivity (reducing power) decreases from left to right.

ii) a) Galvanic cell

$E^\circ_{C^{2+}/C} = -0.76$ V is lower than $E^\circ_{D^{3+}/D^{2+}} = +0.77$ V. So C is oxidised and is the anode. The $\mathrm{D^{3+}/D^{2+}}$ couple is reduced and is the cathode. Both D ions are in solution, so an inert platinum electrode is used at the cathode.

• Anode: $\mathrm{C \longrightarrow C^{2+} + 2e^-}$
• Cathode: $\mathrm{2D^{3+} + 2e^- \longrightarrow 2D^{2+}}$
• Cell reaction: $\mathrm{C + 2D^{3+} \longrightarrow C^{2+} + 2D^{2+}}$

$$\underset{\text{Anode }(-)}{\mathrm{C(s)\,|\,C^{2+}(1\,M)}}\ \Big\|\ \underset{\text{Cathode }(+)}{\mathrm{D^{3+}(1\,M),\,D^{2+}(1\,M)\,|\,Pt}}$$

b) EMF with 1 M solutions

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.77 - (-0.76) = +1.53\ \text{V}$$

c) Will $\mathrm{C^{2+} + 2D^{2+} \rightarrow C + 2D^{3+}}$ occur?

No. (The equation as printed needs the coefficient 2 before $\mathrm{D^{3+}}$ to balance.) In this reaction $\mathrm{C^{2+}}$ would be reduced (cathode) and $\mathrm{D^{2+}}$ oxidised (anode), so

$$E^\circ_{cell} = E^\circ_{C^{2+}/C} - E^\circ_{D^{3+}/D^{2+}} = -0.76 - 0.77 = -1.53\ \text{V}$$

The emf is negative, so $\Delta G^\circ = -nFE^\circ = -2 \times 96500 \times (-1.53) \approx +295\ \text{kJ}$ is positive. The reaction is non-spontaneous. It is the reverse of the cell reaction above: C is a stronger reducing agent than $\mathrm{D^{2+}}$, so $\mathrm{D^{2+}}$ cannot reduce $\mathrm{C^{2+}}$.

a) Standard reduction potential

It is the potential of an electrode, relative to the standard hydrogen electrode (taken as 0.00 V), when the ions are at 1 M concentration (gases at 1 atm) and the temperature is 298 K, written for the reduction process $M^{n+} + ne^- \rightarrow M$.

b) Order of reducing power

The lower (more negative) the reduction potential, the greater the tendency to lose electrons. So

$$A\ (-0.76\ \text{V}) > B\ (+0.34\ \text{V}) > C\ (+0.80\ \text{V})$$

Reducing power: A > B > C.

c) Cell notation (taking B as a divalent metal, $B^{2+}/B$)

B has a positive potential, so it is the cathode and the SHE is the anode:

$$Pt,\ H_2(1\ atm)\,|\,H^+(1\ M)\,||\,B^{2+}(1\ M)\,|\,B$$

d) Cell reaction

Anode: $H_2 \rightarrow 2H^+ + 2e^-$

Cathode: $B^{2+} + 2e^- \rightarrow B$

$$H_2 + B^{2+} \rightarrow 2H^+ + B$$

e) Standard emf

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - 0.00 = +0.34\ \text{V}$$

(d) Ag-vessel

A metal can store $CuSO_4$ solution safely only if it is less reactive than copper, i.e. has a higher reduction potential than $Cu^{2+}/Cu$ (+0.34 V). Ag (+0.80 V) cannot displace Cu. Zn (−0.76 V), Fe (−0.44 V) and Al (−1.66 V) are more reactive and would displace copper, e.g. $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$, corroding the vessel.

(a) Standard hydrogen electrode (SHE). Diagram description, to be drawn and labelled:

• A platinum foil coated with platinum black, sealed in a glass tube.
• Pure $H_2$ gas at 1 atm bubbled over the foil, with a gas inlet and outlet.
• The foil dipping in 1 M $H^+$ solution (1 M HCl) at 298 K.
• A salt bridge or side tube connecting to the other half-cell.

Half reaction: $H_2 \rightleftharpoons 2H^+ + 2e^-$, and $E^\circ = 0.00$ V by convention.

(b) The potential of the SHE is arbitrarily fixed as zero. The potentials of all other electrodes are measured by connecting them to it, so it serves as the reference.

(c) Zinc has the lower reduction potential, so it is the anode (oxidation). Lead is the cathode (reduction).

Cell notation:

$$Zn(s) \,|\, Zn^{2+}(1\,M) \,||\, Pb^{2+}(1\,M) \,|\, Pb(s)$$ $$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.13 - (-0.76) = +0.63 \text{ V}$$