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

Step-by-step 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 - (0.0591)/(2) log (50 * 10^(-3))/(1)`

=
`-(0.0591)/(2) log (5 * 10^(-2))`

= 0.038 V

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