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A hypothetical electrochemical cell is shown below:
$\overset{\ominus}{A}\,|\,A^+(xM)\,||\,B^+(yM)\,|\,\overset{\oplus}{B}$
The emf measured is +0.20 V. The cell reaction is :
$\overset{\ominus}{A}\,|\,A^+(xM)\,||\,B^+(yM)\,|\,\overset{\oplus}{B}$
The emf measured is +0.20 V. The cell reaction is :
A
$A^+ + B \rightarrow A + B^+$
B
$A^+ + e^- \rightarrow A$; $B^+ + e^- \rightarrow B$
C
The cell reaction cannot be predicted
D
$A + B^+ \rightarrow A^+ + B$
Detailed Solution
In the cell notation the anode (negative electrode) is written on the left and the cathode (positive electrode) on the right.
$E_{cell} = +0.20$ V is positive, so $\Delta G = -nFE_{cell}$ is negative and the cell reaction as written is spontaneous.
At the anode (left, oxidation): $A \rightarrow A^+ + e^-$
At the cathode (right, reduction): $B^+ + e^- \rightarrow B$
Adding the two half reactions gives the cell reaction:
$A + B^+ \rightarrow A^+ + B$
$E_{cell} = +0.20$ V is positive, so $\Delta G = -nFE_{cell}$ is negative and the cell reaction as written is spontaneous.
At the anode (left, oxidation): $A \rightarrow A^+ + e^-$
At the cathode (right, reduction): $B^+ + e^- \rightarrow B$
Adding the two half reactions gives the cell reaction:
$A + B^+ \rightarrow A^+ + B$