Question

A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.00 M solution of MSO, and the ight half cell with a 30.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C lf right 410 Which electrode will be positive? What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Answer 1

Explanation:

It is known that for high concentration of $M^{2+}$, reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, $E^{o}_{cell}$ = 0 and the general reaction equation is as follows.

         $M^{2+} + M \rightarrow M + M^{2+}$

                3.00 M        n = 2       30 mM

         E = $0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}$

            = $-\frac{0.0591}{2} log (5 \times 10^{-2})$

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.