I write the half-equations: \begin{align} \ce{Zn &<=> Zn^2+ + 2e^-}& E^\circ &= +0.763~\pu{V}\\ \ce{Cd^2+ + 2e- &<=> Cd}& E^\circ &= +0.403~\pu{V} \end{align}

In order to find the $E_\mathrm{cell}$ is it right to say: $$E_\mathrm{cell} = 0.763~\pu{V} + 0.403~\pu{V} = 1.166~\pu{V}?$$

Or should I say $0.403~\pu{V} - 0.763~\pu{V}$?

  • $\begingroup$ 0.763+0.403 = 1.166V is correct. To use -0.763 V you'd need to flip the reaction with Zn. Both cations can't be reduced obviously... $\endgroup$
    – MaxW
    Feb 5, 2017 at 16:43
  • $\begingroup$ Woops! You have a mistake in the Cd half cell reaction. See Avi's answer. $\endgroup$
    – MaxW
    Feb 5, 2017 at 18:11
  • $\begingroup$ The standard reduction potential of cd is -0.403V not +0.403V. $\endgroup$
    – Mitchell
    Feb 6, 2017 at 9:00

2 Answers 2


I will like to correct you that there is one problem with your question. The correct standard reduction potential is: \begin{align} \ce{Cd^2+ + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}. \end{align} And another thing is, you have used symbol $E^\circ$ for oxidation potential of $\ce{Zn}$ which should not be done. You can use there just $E = 0.7628~\pu{V}$ or in a better way, you can just write it with it's reduction potential value as: \begin{align} \ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}. \end{align} Now coming to your question, for these two reactions, \begin{align} \ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}\tag1\label{Zn}\\ \ce{Cd^2+ + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}\tag2\label{Cd} \end{align} $E_\mathrm{cell}$ of any cell reaction is given as $$E^\circ(\text{cell}) = E^\circ(\text{reduced species}) - E^\circ(\text{oxidized species})$$ or $$E^\circ(\text{cell}) = E^\circ(\text{cathode}) - E^\circ(\text{annode}).$$

Here as $E^\circ$ value of reaction $\eqref{Cd}$ is greater (i.e. towards more positive). Hence $\ce{Cd}$ has greater tendency to accept electrons and get reduced. So the $\ce{Zn}$ is going to give electrons and gets oxidized. Now the half cell reaction will be, \begin{align} \ce{Zn &<=> Zn^2+ + 2e^-}\tag{at anode}\label{anode}\\ \ce{Cd^2+ + 2e- &<=> Cd}.\tag{at cathode}\label{cathode} \end{align}

So according to formula \begin{align} E^\circ(\text{cell}) &= E^\circ(\text{cathode}) - E^\circ(\text{annode})\\ E^\circ(\text{cell}) &= -0.403~\pu{V} - (-0.7628~\pu{V}) = 0.3598~\pu{V}\\ \end{align}

The correct answer is $0.3598~\pu{V}$.


To answer this, lets get some basic things clear.

  1. Reduction potential = - Oxidation potential

  2. Oxidation means losing of electrons and reduction means gain of electrons (OIL RIG)

  3. And finally, Oxidation occurs at anode and reduction occurs at cathode (CR AO)

Standard oxidation potential of Zn is +0.763V and the standard oxidation of potential of Cd is 0.403V.

Since, the standard oxidation potential of Zn is more than that of Cd, Zn electrode will act as anode (where oxidation takes place) and Cd will act as cathode (where reduction takes place).

Ecell = (Eoxidation)anode - (Eoxidation)cathode
      = +0.763 - (0.403)
      = 0.36V


Standard reduction potential of Cd electrode (-0.403V) is greater than that of Zn electrode (-0.763V) so Cd electrode will act as cathode and Zn electrode will act as anode.

Ecell = (Ereduction)cathode - (Ereduction)anode
      = -0.403 - (-0.763)
      = 0.36V

And one more important thing, If only electrode potential is mentioned then that electrode potential is to be assumed the reduction potential by default (not the oxidation potential).

  • 2
    $\begingroup$ The mnemonic I was taught for #3 was "AN OX, RED CAT". Also, just worth mentioning that electrode potentials are much more commonly quoted as reduction potentials - not that it invalidates anything you've said, of course. I'd just be hesitant to use oxidation potentials to calculate anything. Not because it's wrong, but just because it's possible to lead to confusion. $\endgroup$ Feb 5, 2017 at 17:29

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