For the cell reaction,
Zn(s) + $C u^{2 +}$ (0.1M) $\to Z n^{2 +}$ (0.01M) + Cu(s)
If $E^{0}$ is the standard e.m.f. of the cell, then

E = E^{0}

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

E >

E_{0}

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

E <

E_{0}

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

None of the above.

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

Open in App

Solution

The correct option is B E > $E_{0}$
$E = E^{0} \frac{- 0.059}{2} l o g [\frac{Z n^{2 +}}{C u^{2 +}}] = E^{0} - \frac{0.059}{2} l o g \frac{0.01}{0.1} = E^{0} - \frac{0.059}{2} l o g \frac{0.01}{0.1} = E^{0} - \frac{0.059}{2} l o g (\frac{1}{10})$
or, E=E^0 + 0.295 $∴$ E > $E^{0}$

Suggest Corrections

Similar questions

For the cell reaction:

$Z n (s) + C u^{2 +} (0.1 M) \to Z n^{2 +} (0.01 M) + C u (s)$

If $E^{0}$ is the standard emf of the cell at 298 K, then:

E^{\circ}

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = 0.76 V

C u \to C u^{2 +} + 2 e^{-}; E^{\circ} = 0.34 V

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

Z n + C u_{(0.1 M)}^{2 +} \to Z n_{(1 M)}^{2 +} + C u

E_{c e l l}^{⊖} = 1.10

E_{C e l l}^{⊖}

E

E_{0}

Z n | Z n^{2 +} | | C u^{2 +} | C u

E_{z n}^{⊖} = 0.736 V; E_{(c u)}^{⊖} = - 0.35 V

Join BYJU'S Learning Program