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Q.

For the process 4OH^–_(aq) → 2H₂O_(l) + O_2(g) + 4e^– taking place at 298K at anode, ΔH is [ Δ_fH of H₂O_(l) and OH^–_(aq) are –286 and –230 KJmol^-1]

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a

+348 kJ

b

-348 kJ

c

-174 kJ

d

+87 kJ

answer is A.

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Detailed Solution

$\begin{gathered} {\Delta _r}(H) = \sum \Delta H_{products}^o - \sum \Delta H_{reactants}^o \\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = [2{\Delta _f}H({H_2}O) + {\Delta _f}H({O_2})] - [4{\Delta _f}H(O{H^ - })] \\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = [2( - 286) + 0] - [4( - 230)] \\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 348KJ\\ \end{gathered}$

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For the process 4OH–(aq) → 2H2O(l) + O2(g) + 4e– taking place at 298K at anode, ΔH is [ ΔfH of H2O(l) and OH–(aq) are –286 and –230 KJmol-1]