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Question

For the following electrochemical cell at 298 K,
Pt(s)|H2(g,1bar)|H+(aq,1M)||M4+(aq),M2+(aq)|Pt(s)
Ecell=0.092V when[M2++(aq)][M4+(aq)]=10x
Given:EM4+/M2+=0.151V;2.303RTF=0.059V
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
Cell reaction
M4++H2M2++2HEcell=Ecell0.062log[M2+][M4+]0.092=0.1510.03logM2+M4+=0.06=0.03log10[M+2][M+4][M2+][M4+]=102 [GivenM2+(aq)[M4+(aq)]]10x=102=x=2

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