Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins

Explanation:

We have to start with the half-reactions for both ions:

V= -0.76

V= +0.80

If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:

V= +0.76

V= +0.80

If we want to calculate ºE we have to add the two values, so:

ºE=0.76+0.80 = 1.56 V

Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:

On this case, Q is equal to:

Because the total reaction is:

So, the value of "Q" is:

Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:

I hope it helps!