A) Increase by 0.059 V
B) Decrease by 0.059 V
C) Increase by 0.41 V
D) Decrease by 0.41 V
Correct Answer: D
Solution :
\[2{{H}^{+}}+2{{e}^{-}}\xrightarrow{\,}\,{{H}_{2}}\] \[{{E}_{({{H}^{+}}/{{H}_{2}})}}={{E}_{({{H}^{+}}/{{H}_{2}})}}\circ \,-\,\frac{0.059}{2}\,\log \,\frac{1}{{{[{{H}^{+}}]}^{2}}}\] \[=0+\frac{0.059}{2}\,\log \,[{{H}^{+}}]\,=-0.059\,pH\] when \[pH=0\] \[{{E}_{{{H}^{+}}/{{H}_{2}}}}=0\] when \[pH=0\] \[{{E}_{{{H}^{+}}/{{H}_{2}}}}=-0.059\times 7\] = - 0.41 V i.e., reduction potential decreases by 0.41 V.
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