Question
The voltage of the cell given below is −0.61 V. Pt|H2 (1 bar)|NaHSO3 (0.4 M), Na2SO3 (6.4 × 10−2 M)||Zn2+ (0.4 M)|Zn If $E_{Zn^{2 +}|Zn}^{o}$= − 0.76 V, Calculate Ka2 of H2SO4. (2.303 RT/F = 0.06)
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Solution
Net cell reaction may be written as
$H_{2} + Zn^{2 +} \rightleftharpoons 2H^{+} + Zn $$${E_{cell} = E_{cell}^{o} - \frac{0.06}{2}.\log\frac{\left\lbrack H^{+} \right\rbrack^{2}}{P_{H_{2}}.\left\lbrack Zn^{2 +} \right\rbrack} }{\text{Or, }( - 0.61) = ( - 0.76) - \frac{0.06}{2}.\log\frac{\left\lbrack H^{+} \right\rbrack^{2}}{1 \times 0.4} }{\therefore\left\lbrack H^{+} \right\rbrack = 2 \times 10^{- 3}\text{ M} }{\text{Now, }K_{a_{2}} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack SO_{3}^{2 -} \right\rbrack}{\left\lbrack HSO_{3}^{-} \right\rbrack} = \frac{\left( 2 \times 10^{- 3} \right) \times \left( 6.4 \times 10^{- 2} \right)}{0.4} = 3.2 \times 10^{- 4}}$$
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