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To calculate the electrode potential of the zinc rod dipped in a 0.1 M solution of ZnSO₄, we can use the Nernst equation:
E=E0−0.0592nlogQE=E0−n0.0592logQ
Where:
The half-reaction for the zinc electrode can be written as:
Zn2++2e−→ZnZn2++2e−→Zn
Given that the solution is 95% dissociated, the concentration of Zn2+Zn2+ ions will be 0.1 M * 0.95 = 0.095 M.
The reaction quotient, QQ, is given by:
Q=[Zn][Zn2+]=1[Zn2+]Q=[Zn2+][Zn]=[Zn2+]1
Since the stoichiometric coefficient of Zn2+Zn2+ in the half-reaction is 1, and the stoichiometric coefficient of ZnZn is also 1, the concentration of Zn2+Zn2+ ions is equal to the concentration of ZnZn atoms.
Q=10.095=10.526Q=0.0951=10.526
The standard electrode potential (E0E0) for the zinc electrode is given as -0.76 V.
The number of moles of electrons transferred (nn) is 2 (as per the balanced half-reaction).
Now, plug in these values into the Nernst equation:
E=−0.76−0.05922log(10.526)E=−0.76−20.0592log(10.526)
Now, calculate the electrode potential EE:
E=−0.76−0.05922×1.022E=−0.76−20.0592×1.022
E=−0.76−0.0302E=−0.76−0.0302
E=−0.7902E=−0.7902
So, the electrode potential is approximately -0.7902 V.
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