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Question

The equilibrium constant for the redox change
$$[2Ag(NH_3)_2^+ + C_6H_{12} O_6 + H_2O  \rightarrow C_6H_{12} O_7 + 2Ag(s) + C_6H_{12} O_2 + 2H^+]$$ is:


A
5.15×109
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B
4,2×105
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C
1.72×1011
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D
4.2×105
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Solution

The correct option is D $$5.15 \times 10^9$$
$$[2Ag(NH_3)_2^+ + C_6H_{12} O_6 + H_2O  \rightarrow C_6H_{12} O_7 + 2Ag(s) + C_6H_{12} O_2 + 2H^+]$$
$$E = E^o_{OP_{Glucose}} + E^o_{RP_{Ag(NH_3)_2^+ / Ag}} = - 0.05 + 0.337 = 0.287 V$$
According to Nernst equation:
$$E = E^o-0.0591logQ/n$$
And at eqm. $$E=0$$
so 
$$\displaystyle E^o = \frac{2.303 RT}{nF} log K = \frac{0.0591}{2} log  K$$
$$\therefore log K = \displaystyle \frac{2 \times 0.287}{0.0591} = 9.71$$
$$K = 5.15 \times 10^9$$

Chemistry

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