Consider the following redox reaction: $$MnO_4^- + H^+ + H_2C_2O_4 \rightleftharpoons Mn^{2+} + H_2O + CO_2$$
The standard reduction potentials are given as below:
$$E^0_{MnO_4^-/Mn^{2+}} = +1.51$$ V; $$E^0_{CO_2/H_2C_2O_4} = -0.49$$ V
If the equilibrium constant of the above reaction is given as $$K_{eq} = 10^x$$, then the value of $$x$$ = ______ (nearest integer)

• Reduction half-reaction:
  $$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad (\text{Electrons gained} = 5)$$
• Oxidation half-reaction:
  $$\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^- \quad (\text{Electrons lost} = 2)$$

To balance the overall redox equation, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.

• Total net electrons transferred ($$n$$): $$5 \times 2 = \mathbf{10}$$

2. Calculate Standard Cell Potential ($$E^0_{\text{cell}}$$)

Using the standard reduction potentials provided:

• $$E^0_{\text{cathode}} (\text{Reduction of MnO}_4^-) = +1.51\text{ V}$$
• $$E^0_{\text{anode}} (\text{Reduction of CO}_2) = -0.49\text{ V}$$

$$E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}$$

$$E^0_{\text{cell}} = 1.51\text{ V} - (-0.49\text{ V}) = \mathbf{2.00\text{ V}}$$

3. Connect $$E^0_{\text{cell}}$$ to the Equilibrium Constant ($$K_{eq}$$)

Using the Nernst equation relationship at equilibrium ($$298\text{ K}$$):

$$E^0_{\text{cell}} = \frac{0.0591}{n} \log_{10}(K_{eq})$$

Substitute the values $$E^0_{\text{cell}} = 2.00$$ and $$n = 10$$:

$$2.00 = \frac{0.0591}{10} \log_{10}(K_{eq})$$

$$20.0 = 0.0591 \log_{10}(K_{eq})$$

$$\log_{10}(K_{eq}) = \frac{20.0}{0.0591} \approx 338.4$$

x = 338