Question

(i) Calculate \( E^\circ_{cell} \) for the following reaction which is at equilibrium :

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

Equilibrium constant \( (K_c) \) for the cell is \( 10^{15} \).

[Given : log 10 = 1]

(ii) Write anode, cathode and overall reaction of lead storage battery when it is in use.

(iii) How much electricity is required in coulombs for the oxidation of 1 mol of FeO to \( \mathrm{Fe_2O_3} \) ?

Hint:

Use the relation between the standard cell potential and log K with n = 2. For FeO to Fe2O3, one electron is lost per iron atom, so the charge is one faraday.

Answer 1

Step 1: Understanding the Question:
Part (i) links the equilibrium constant to the standard cell potential. Part (ii) asks for the reactions of the lead storage battery. Part (iii) is a Faraday law calculation.

Step 2: Part (i), Formula:
At equilibrium the actual cell potential is zero. The standard potential is linked to \( K_c \) by the relation below at 298 K.
\[ E^\circ_{cell} = \frac{0.0591}{n} \log K_c \]
In the reaction, copper loses 2 electrons. So \( n = 2 \).

Step 3: Part (i), Calculation:
Given \( K_c = 10^{15} \), so \( \log K_c = 15 \log 10 = 15 \).
\[ E^\circ_{cell} = \frac{0.0591}{2} \times 15 = 0.44325\ \mathrm{V} \approx 0.44\ \mathrm{V} \]

Step 4: Part (ii), Lead storage battery on discharge:
Lead is the anode and lead dioxide is the cathode. The electrolyte is about 38 percent \( \mathrm{H_2SO_4} \).
Anode (oxidation):
\[ \mathrm{Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-} \]
Cathode (reduction):
\[ \mathrm{PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)} \]
Overall reaction:
\[ \mathrm{Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)} \]

Step 5: Part (iii), Charge for FeO to Fe2O3:
In FeO, iron is present as \( \mathrm{Fe^{2+}} \). In \( \mathrm{Fe_2O_3} \), it is \( \mathrm{Fe^{3+}} \).
\[ \mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-} \]
So one mole of FeO gives one mole of electrons. The charge on one mole of electrons is one faraday.
\[ Q = 1\ \mathrm{mol} \times 96500\ \mathrm{C\ mol^{-1}} = 96500\ \mathrm{C} \]

Final Answer:
(i) \( E^\circ_{cell} \) is about 0.44 V. (ii) Reactions are given in step 4. (iii) The charge needed is 96500 C.
\[ \boxed{E^\circ_{cell} = 0.44\ \mathrm{V};\ Q = 96500\ \mathrm{C}} \]