Question

A Zn wire and Ag/AgCl reference electrode (E° = 0.197 V) are placed into a solution of ZnSO4. The Zn wire is attached to the positive terminal and the Ag/AgCl electrode is attached to the negative terminal of the potentiometer. Calculate the [Zn2+ ] in the solution if the cell potential, Ecell, is -1.061 V. The standard reduction potential of the Zn2+ /Zn half-reaction is –0.762 V.

[Zn2+]= ? M

Answer 1

Answer on Question #71801, Chemistry / General Chemistry :

A Zn wire and Ag/AgCl reference electrode ( $E^{*} = 0.197\mathrm{V}$ ) are placed into a solution of $\mathrm{ZnSO_4}$ . The Zn wire is attached to the positive terminal and the Ag/AgCl electrode is attached to the negative terminal of the potentiometer. Calculate the $[\mathrm{Zn}^{2+}]$ in the solution if the cell potential, Ecell, is -1.061 V. The standard reduction potential of the $\mathrm{Zn}^{2+}/\mathrm{Zn}$ half-reaction is -0.762 V.

$$[ \mathrm {Z n} ^ {2 +} ] =? \mathrm {M}.$$

Solution.

$$E ^ {0} = 0. 1 9 7 V$$ $$E = 1. 0 6 1 V$$ $$E \left(Z n ^ {2 +} / Z n\right) = - 0. 7 6 2 V$$ $$\left[ Z n ^ {2 +} \right] -?$$

The cell potential, Ecell, is -1.061 V:

$$E = 1. 0 6 1 V = E ^ {0} - E \left(Z n ^ {2 +} / Z n\right)$$

And:

$$1. 0 6 1 V = 0. 1 9 7 V - \left(E ^ {0} \left(Z n ^ {2 +} / Z n\right) + \frac {0 . 0 5 9}{2} \cdot \lg \left[ Z n ^ {2 +} \right]\right)$$ $$1. 0 6 1 V = 0. 1 9 7 V - \left(- 0. 7 6 2 + \frac {0 . 0 5 9}{2} \cdot \lg \left[ Z n ^ {2 +} \right]\right)$$ $$\frac {0 . 0 5 9}{2} \cdot \lg \left[ Z n ^ {2 +} \right] = 0. 1 9 7 V + 0. 7 6 2 V - 1. 0 6 1 V$$ $$\lg \left[ Z n ^ {2 +} \right] = - 3. 4 5 7$$ $$\left[ Z n ^ {2 +} \right] = 1 0 ^ {- 3. 4 5 7} = 3. 4 8 6 \cdot 1 0 ^ {- 4} M$$ $$\left[ Z n ^ {2 +} \right] = 3. 4 8 6 \cdot 1 0 ^ {- 4} M$$

Answer: $\left[Z n^{2+}\right] = 3.486 \cdot 10^{-4} M$ .

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