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NEET CHEMISTRYEasy

A hypothetical electrochemical cell is shown below: AA+(x M)B+(y M)B\text{A} | \text{A}^+(x\text{ M}) || \text{B}^+(y\text{ M}) | \text{B} The EMF measured is +0.20 V+0.20 \text{ V}. The cell reaction is:

A

A++BA+B+\text{A}^+ + \text{B} \rightarrow \text{A} + \text{B}^+

B

A++eA;B++eB\text{A}^+ + e^- \rightarrow \text{A} ; \text{B}^+ + e^- \rightarrow \text{B}

C

the cell reaction cannot be predicted

D

A+B+A++B\text{A} + \text{B}^+ \rightarrow \text{A}^+ + \text{B}

Step-by-Step Solution

According to the IUPAC convention for representing an electrochemical cell, the anode (where oxidation occurs) is written on the left, and the cathode (where reduction occurs) is written on the right. For the given cell AA+(x M)B+(y M)B\text{A} | \text{A}^+(x\text{ M}) || \text{B}^+(y\text{ M}) | \text{B}: Oxidation half-reaction at the anode (left): AA++e\text{A} \rightarrow \text{A}^+ + e^- Reduction half-reaction at the cathode (right): B++eB\text{B}^+ + e^- \rightarrow \text{B} Adding the two half-reactions gives the overall cell reaction: A+B+A++B\text{A} + \text{B}^+ \rightarrow \text{A}^+ + \text{B} The positive value of the EMF (+0.20 V+0.20 \text{ V}) indicates that the cell reaction is spontaneous in the forward direction as written.

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