You visited us 1 times! Enjoying our articles? Unlock Full Access!

Nernst Equation

If excess of ...

Question

If excess of $Z n$ is added to $1.0 M$ solution of $C u S O_{4}$ , find the concentration of $C u^{2 +}$ ions at equilibrium.
Given: $E_{(Z n^{2 +} | Z n)}^{0} = - 0.76 V, E_{(C u^{2 +} C u)}^{0} = - 0.34 V$

5 \times 10^{- 25}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 \times 10^{- 17}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 \times 10^{- 38}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 \times 10^{- 9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Given $Z n ∣ ∣ {Z n}^{2 +} (0.001 M) ∣ ∣ ∣ ∣ {C u}^{2 +} (0.1 M) ∣ ∣ C u$
Overall cell reaction:
$Z n ⟶ {Z n}^{2 +} + 2 e^{-}$
${C u}^{2 +} + 2 e^{-} ⟶ C u$
$Z n + {C u}^{2 +} ⟶ {Z n}^{2 +} + C u$
$E_{c e l l}^{0}$ = Standard reduction potential of cathode + standard oxidation potential of anode
$E_{c e l l}^{0} = 0.34 V + 0.76 V$
$E_{c e l l}^{0} = 1.1 V$
$K_{C} = \frac{[{Z n}^{2 +}]}{[{C u}^{2 +}]} = \frac{10^{- 3}}{10^{- 1}} = 10^{- 2}$
EMF of the cell at any electrode concentration is:
$E = E^{0} - \frac{0.059}{n} l o g (K_{C})$
$= 1.1 - \frac{0.059}{2} l o g (10^{- 2})$
$= 1.1 - \frac{0.059}{2} \times (2) = 1.1 - 0.059$
$= 1.041 V$

Suggest Corrections

Similar questions

Z n

1.0 M

C u S O_{4}

C u^{2 +}

E_{(Z n^{2 +} | Z n)}^{0} = - 0.76 V, E_{(C u^{2 +} C u)}^{0} = - 0.34 V

E^{\circ}

C u_{(s)} + Z n_{(a q)}^{2 +} \to Z n_{(s)} + C u_{(a q)}^{2 +}

E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{Z n^{2 +} / Z n}^{0} = - 0.76 V

Z n (s) / Z n^{2 +} (a q) ∥ C u^{2 +} (a q) / C u (s)

{E^{0}}_{{Z n}^{2 +} / Z n} =

-

{E^{0}}_{{C u}^{2 +} / C u} =

+

{E^{0}}_{c e l l}

K

Z n

1.0 M

C u S O_{4}

C u^{2 +}

E_{(Z n^{2 +} | Z n)}^{0} = - 0.76 V, E_{(C u^{2 +} C u)}^{0} = - 0.34 V

Z n

1.0 M

C u S O_{4}

C u^{2 +}

E_{(Z n^{2 +} | Z n)}^{0} = - 0.76 V, E_{(C u^{2 +} C u)}^{0} = - 0.34 V

Join BYJU'S Learning Program