what is the calculated value of the cell potential at 298k for an electrochemical cell with the following reaction, when the cu2+ concentration is 2.32×10-4 m and the mn2+ concentration is 1.04 m ?
cu2+(aq) + mn(s) cu(s) + mn2+(aq)

the cell reaction as written above is spontaneous for the concentrations given: (true/false)

The correct answer is true.

Explanation:

The given cell is:

Half reactions for the given cell follows:

Oxidation half reaction:

Reduction half reaction:

Net reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

where,

= electrode potential of the cell = ?V

= standard electrode potential of the cell = +1.52 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

As, the EMF of the cell is coming out to be positive. Thus, the cell reaction is spontaneous for the given concentration.

Hence, the correct answer is true.