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Question

For the following electrochemical cell at 298 K,
Pt(s)|H2(g,1 bar)|H+(aq,1M)||M4+(aq),M2+(aq)|Pt(s)Ecell=0.092 V when [M2+(aq)][M4+(aq)]=10xGiven:E0M4+/M2+=0.151 V;2.303RTF=0.059 V
The value of x is:

A
-2
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B
-1
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C
1
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D
2
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Solution

The correct option is D 2
Anode:H22H++2eCathode:M4++2eM2+
Net cell reaction: H2+M4+2H++M2+Ecell=E0cell0.0592log[H+][M2+][M4+]×PH20.092=0.1510.0592log 10x0.092=0.1510.0592x0.059 x2=0.1510.0920.059 x=0.059×2x=2

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