Question

calculate the potential of the cell consist of a silver electrode dipping in silver nitrate solution with aAg+ =0.0100M @nd a SCE reference electrode.​

Answer 1

Given that,

A silver electrode dipping in silver nitrate solution with a 0.25M.

Suppose, A zinc electrode dipping in zinc nitrate solution with a 0.010M.

For silver,

$E^{\circ}_{cathode}=0.80\ V$

For zinc,

$E^{\circ}_{anode}=+0.763\ V$

We need to calculate the value of $E^{\circ}_{cell}$

Using formula of $E^{\circ}_{cell}$

$E^{\circ}_{cell}=E^{\circ}_{cathode}+[tex]E^{\circ}_{anode}$

Put the value into the formula

$E^{\circ}_{cell}=0.80+0.763$

$E^{\circ}_{cell}=1.56\ V$

We need to calculate the potential of the cell

Using formula of potential

$E_{cell}=E^{\circ}_{cell}-\dfrac{0.0257}{n}ln(\dfrac{[Zn^{2+}]}{[Ag^{+}]^2})$

Put the value into the formula

$E_{cell}=1.56-\dfrac{0.0257}{2}ln(\dfrac{0.010}{(0.25)^2})$

$E_{cell}=1.58\ V$

Hence, The potential of the cell is 1.58 V.