Electro Chemistry Part 3

Concept:The cell potential depends on the concentration of hydroxide ions, which is determined by the pH of the buffer. The buffer pH is calculated using the Henderson–Hasselbalch equation for the $\mathrm{H_2CO_3/HCO_3^-}$ pair.Explanation:Step 1: Identify cathode and anode.The half-reaction with the higher standard potential acts as cathode. Here, $\mathrm{Bi_2O_3/Bi}$ has $E^\circ = -0.44\ \mathrm{V}$ and $\mathrm{Sn(OH)_6^{2-}/HSnO_2^-}$ has $E^\circ = -0.90\ \mathrm{V}$. So Bi couple is cathode, Sn couple is anode.Step 2: Write Nernst equations for each half-cell.For Sn couple (reduction):$\mathrm{Sn(OH)_6^{2-} + 2e^- \rightarrow HSnO_2^- + 3OH^- + H_2O}$$E_{\mathrm{Sn}} = -0.90 - \frac{0.059}{2} \log\frac{[\mathrm{HSnO_2^-}][\mathrm{OH^-}]^3}{[\mathrm{Sn(OH)_6^{2-}}]}$Given $[\mathrm{Sn(OH)_6^{2-}}] = 0.5\ \mathrm{M}$, $[\mathrm{HSnO_2^-}] = 0.05\ \mathrm{M}$, ratio = $0.1$.For Bi couple (reduction):$\mathrm{Bi_2O_3(s) + 3H_2O + 6e^- \rightarrow 2Bi(s) + 6OH^-}$$E_{\mathrm{Bi}} = -0.44 - \frac{0.059}{6} \log[\mathrm{OH^-}]^6 = -0.44 - 0.059 \log[\mathrm{OH^-}]$Step 3: Combine to get cell potential.$E_{\mathrm{cell}} = E_{\mathrm{Bi}} - E_{\mathrm{Sn}}$Substitute and simplify:$E_{\mathrm{cell}} = [-0.44 - 0.059 \log[\mathrm{OH^-}]] - \left[-0.90 - \frac{0.059}{2} \log(0.1 \times [\mathrm{OH^-}]^3)\right]$$= 0.46 - 0.059 \log[\mathrm{OH^-}] + \frac{0.059}{2} (\log 0.1 + 3 \log[\mathrm{OH^-}])$Since $\log 0.1 = -1$:$= 0.46 + \frac{0.059}{2}(-1) + \left(-0.059 + \frac{3 \times 0.059}{2}\right) \log[\mathrm{OH^-}]$$= 0.4305 + 0.0295 \log[\mathrm{OH^-}]$Step 4: Use given cell potential to find $\mathrm{pOH}$.Given $E_{\mathrm{cell}} = 235.3\ \mathrm{mV} = 0.2353\ \mathrm{V}$:$0.2353 = 0.4305 + 0.0295 \log[\mathrm{OH^-}]$$\log[\mathrm{OH^-}] = \frac{0.2353 - 0.4305}{0.0295} = \frac{-0.1952}{0.0295} \approx -6.6$Thus $\mathrm{pOH} \approx 6.6$, so $\mathrm{pH} = 14 - 6.6 = 7.4$.Step 5: Use buffer equation to find the required volume $x$.Buffer: $\mathrm{H_2CO_3}$ (weak acid) and $\mathrm{HCO_3^-}$ (conjugate base).Henderson–Hasselbalch: $\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]}$$7.4 = 6.11 + \log\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]}$$\log\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]} = 1.29$$\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]} = \text{antilog}(1.29) = 19.5$Moles of $\mathrm{H_2CO_3}$: $2\ \mathrm{M} \times 10\ \mathrm{mL} = 0.020\ \text{mol}$Moles of $\mathrm{HCO_3^-}$: $5\ \mathrm{M} \times x\ \mathrm{mL} = 0.005x\ \text{mol}$Ratio: $\frac{0.005x}{0.020} = 19.5$$0.005x = 0.39 \quad \Rightarrow \quad x = 78$Answer:$x = 78\ \text{mL}$ (nearest integer). Option A is correct.