A) 0.18 V
B) 0.28 V
C) 0.38 V
D) 0.48 V
Correct Answer: C
Solution :
\[\frac{1}{2}{{H}_{2}}\to {{H}^{+}}({{10}^{-8}}M)+{{e}^{-}}(\text{oxidation})\] \[{{H}^{+}}(0.025\,M)+{{e}^{-}}\to \frac{1}{2}{{H}_{2}}(\text{reduction})\] Cell reaction is : \[{{H}^{+}}(0.025\,M)\to {{H}^{+}}({{10}^{-8}}\,M)\]; \[{{E}_{\text{cell}}}=0.38\,\,V\].
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