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Question

For the half cell reaction at 298 K,
H+(aq)+e12 H2(g) (1 atm)

A
EH+/H2=59.2 ×pH mV
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B
E0H+/H2=59.2 ×pH mV
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C
E0H+/H2=pH×log[H+] V
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D
EH+/H2=pH×log[H+] V
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Solution

The correct option is A EH+/H2=59.2 ×pH mV
H+(aq)+e12 H2(g)
EH+/H2=E0H+/H20.05921log(pH2)12[H+]=0+0.0592×log [H+]=0.0592×pH V=59.2×pH mV

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