Question

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

Answer 1

`1.66~V`

Step-by-step explanation:

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

`Zn^+^2~+2e^-~->Zn` V= -0.76

`Ag^+~e^-~->~Ag` V= +0.80

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

`Zn~->~Zn^+^2~+2e^-~` V= +0.76

`Ag^+~+~e^-~->~Ag` 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:

`E=ºE~+~(0.059)/(n)LogQ`

On this case, Q is equal to:

`Q=([Zn^+^2])/([Ag^+]^2)`

Because the total reaction is:

`Zn~+~2Ag^+~->~Zn^+^2~+~2Ag`

So, the value of "Q" is:

`Q=([0.052 M])/([0.0042]^2)=2947.84`

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

`E=1.56~V~+~(0.059)/(2)Log(2947.84)=1.66~V`

I hope it helps!