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

EMF

The E 0 in ...

Question

The $E^{0}$ in the given diagram is :-

0.5

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

0.6

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

0.7

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

0.8

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

Open in App

Solution

The correct option is B 0.6
We have,
$n F E_{c e l l}^{0} = n F E_{1}^{0} + n F E_{2}^{0}$
where n = change in number of electron
$6 E^{0} = 4 * 0.54 + 0.71 * 2$
$E^{0} = 0.59$

Suggest Corrections

Similar questions

E (E > E_{0})

E_{0}

K^{+} + e^{-} \to K; E^{0} = - 2.93 V A g^{+} + e^{-} \to A g; E^{0} = 0.80 V H g_{2}^{2 +} + e^{-} \to 2 H g; E^{0} = 0.79 V M g^{2 +} + e^{-} \to M g; E^{0} = - 2.37 V C r^{3 +} + e^{-} \to C r; E^{0} = - 0.74 V

M n^{2 +} + 2^{e -} \to M n; E^{0} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 V

E^{0}

3 M n^{2 +} \to M n + 2 M n^{3 +}

N O_{3}^{-}

N O_{3}^{-}

E^{0}

+ 0.96 V

E^{0}

V^{2 +} (a q) + 2 e^{-} \to V (s); E^{0} = - 1.19 V

F e^{3 +} (a q) + 3 e^{-} \to F e (s); E^{0} = - 0.04 V

A u^{3 +} (a q) + 3 e^{-} \to A u (s); E^{0} = + 1.40 V

H g^{2 +} (a q) + 2 e^{-} \to H g (s); E^{0} = + 0.86 V

E^{0}

I_{2} (g) + 2 e^{-} \to 2 I^{-} (a q); E^{0} = + 0.54 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{0} = + 1.36 V

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = + 0.76 V

C e^{4 +} (a q) + e^{-} \to C e^{3 +} (a q); E^{0} = + 1.60 V

S n^{4 +} (a q) + 2 e^{-} \to S n^{2 +} (a q); E^{0} = + 0.15 V

Join BYJU'S Learning Program