Calculate E naught cell for the reaction b
5Cd(s) + 2MnO4-(aq) + 16H+(aq) ==> 5Cd^2+(aq) = 2Mn^2+(aq) + 8H2O(l) given the following standard reduction potentials
MnO4^-(aq) + 8H^+(aq) = 5e^- ==> Mn^2+(aq) + 4 H2O(l) E naught = +1.52V
Cd^2+(aq) + 2e^- ==> Cd(s) E naught = -0.40V
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5.04
To calculate the standard cell potential, E°cell, for the given reaction, you need to use the standard reduction potentials of the half-reactions involved.
The given reaction consists of two half-reactions:
1) MnO4^-(aq) + 8H^+(aq) + 5e^- → Mn^2+(aq) + 4H2O(l) (E° = +1.52V)
2) Cd^2+(aq) + 2e^- → Cd(s) (E° = -0.40V)
To calculate E°cell, you find the difference between the reduction potentials of the two half-reactions:
E°cell = E°cathode - E°anode
In this case, the reduction potential of the cathode reaction (the reduction that occurs at the cathode) is +1.52V, and the reduction potential of the anode reaction (the oxidation that occurs at the anode) is -0.40V.
Substituting these values into the equation, we get:
E°cell = (+1.52V) - (-0.40V)
= +1.52V + 0.40V
= +1.92V
Therefore, the standard cell potential, E°cell, for the given reaction is +1.92V.