For the following electrochemical cell at 298 K, Pt(s)|H2(g,1bar)|H+(aq,1M)||M4+(aq),M2+(aq)|Pt(s) Ecell=0.092Vwhen[M2++(aq)][M4+(aq)]=10x Given:E∘M4+/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++H2→M2++2H⊕Ecell=E∘cell−0.062log[M2+][M4+]0.092=0.151−0.03logM2+M4+=−0.06=−0.03log10[M+2][M+4][M2+][M4+]=102[GivenM2+(aq)[M4+(aq)]]10x=102=x=2