Oxidation (anode): H_{2}(g) ==> 2H^{+} + 2e-

Reduction (cathode): Cd^{2+}(aq) + 2e- ==> Cd(s)

Overall redox: H_{2}(g) + Cd^{2+}(aq) ==> 2H^{+}(aq) + Cd(s)

Eºcell = cathode - anode = -0.403 - 0.00 = -0.403 V

Using the Nernst equation: Ecell = Eºcell - 0.0592/n log [H+]^{2}/(PH_{2})[Cd^{2+}]

Now, I'm going to use the value you gave for Ecell of 0.3629 V, but I believe it should be -0.3629. If my assumption is correct, you will have to go back and re-do the calculations plugging in the negative value where I have the positive 0.3629.

0.3629 = -0.403 - 0.0592/2 log [H+]^{2} / (0.707)(1)

0.0592/2 log [H^{+}]^{2} / 0.707 = -0.403 - 0.3629

0.0296 log [H^{+}]^{2} /0.707 = -0.7659

0.0296 log [H^{+}]^{2} = -0.5415

2 log [H^{+}] = -18.3

log [H^{+}] = -9.15

[H^{+}] = 7.1x10^{-10} M (which doesn't make sense, so my assumption that the original Ecell should be negative was probably correct, and you can now re-do the calculations using the negative value).

J.R. S.

7d

Nahida S.

yes Ecell is negative13d