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saucer5

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a) Calculate the potential of an electrode consisting of a strip of Zn in contact with a solution containing 0.01 mol Zn2+ and 3.0 mol CN‐ per litre. The ε° value for aquated Zn2+ is ‐0.76 V, and β = 1 x 10^9 for the Zn2+ / CN‐ equilibrium.

b) ε = ε° ‐ (RT/nF) ln Q

where Q is the reaction quotient. Also useful is that β= [Zn(CN)]/[CN]^2[Zn]

n is the number of electrons transferred to transform Zn ion to Zn solid (2)

c) I tried to do a RICE table using β = 10^9 = x / (0.01-x) (3-2x)^2. assuming that on the bottom, the x values are negligible. However, this isn't correct.

I am given that ε = -0.76, RT/nF = 0.0128. The main problem in this computation is that I have no idea what the reaction quotient should be. I know it is usually based on the half cell equation, which would be ln (1/Zn2+), which I thought I could figure out using the RICE table...but given that the x is negligible...arhghweoufwe!

Help!