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Question

# For the cell, Zn|Zn2+(1 M)||H+(1 M)|Pt(H2, 1 bar),E∘cell=0.74 V and for the cell Pt(H2)|H+(1 M)||Ag+(1 M)|Ag,E∘cell=0.80 V Thus, the value of Ecell and spontaneity for the following reaction respectively are Ag|Ag+(0.1 M)||Zn2+(0.1 M)|Zn

A
1.44 V and spontaneous
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B
0.4 V and spontaneous
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C
1.44 V and non-spontaneous
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D
1.53 V and non-spontaneous
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Solution

## The correct option is D −1.53 V and non-spontaneousFor the cell, Zn|Zn2+(1 M)||H+(1 M)|Pt(H2, 1 bar) E∘cell=E∘SHE−E∘Zn2+/Zn 0.74=0.00−E∘Zn2+/Zn ∴E∘Zn2+/Zn=−0.74 V Similarly, we can calculate the value of E∘Ag+/Ag: Pt(H2,1 bar)|H+(1M)||Ag+(1M)|Ag E∘cell=E∘Ag+/Ag−E∘SHE ⇒E∘Ag+/Ag=0.80 V Now, for the cell Ag|Ag+(0.1 M)||Zn2+(0.1 M)|Zn E∘cell=E∘Zn2+/Zn−E∘Ag+/Ag=−0.74 V−(0.80 V)=−1.54 V Reaction quotient(Q)=[Ag+]2[Zn2+]=(0.1)2(0.1)=0.010.1=110 ∴Ecell=E∘cell−2.303RTnFlogQ At, 298 K,2.303RTF=0.0591 V ∴Ecell=−1.54−0.05912log(110) =−1.54+0.05912log10 =−1.54+0.0295 =−1.51 V Ecell<0, hence reaction is non-spontaneous. So, (d) is correct.

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