For an electrochemical cell Sn s - Sn 2- aq- 1 M P b 2- aq- 1 M - Pb s the ratio - Sn 2- Pb 2- when this cell attains equilibrium is - Given- E - Sn 2- Sn -0-14 V - E Pb 2- Pb -0-13 V - 2-303 RT - F -0-06 V

Anodic half: Sn → Sn^2+ + 2e^ Cathodic half: Pb^2+ + 2e^ → Pb Net reaction: Sn + Pb^2+→ Pb+Sn^2+ E_cell^∘ = E_cathode^∘ E_anode^∘ E_cell^∘ = 0.01 V E_cell = ...

You visited us 5 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

New Capacitance in Dielectric

For an electr...

Question

For an electrochemical cell $S n (s) | S n^{2 +} (a q ., 1 M) | | P b^{2 +} (a q ., 1 M) | P b (s)$ the ratio $\frac{[S n^{2 +}]}{[P b^{2 +}]}$ when this cell attains equilibrium is . (Given: $E_{\frac{S n^{2 +}}{S n}}^{\circ} = - 0.14 V; E_{\frac{P b^{2 +}}{P b}}^{\circ} = - 0.13 V, \frac{2.303 R T}{F} = 0.06 V)$ .

Open in App

Solution

Anodic half: Sn $\to S n^{2 +} + 2 e^{-}$
Cathodic half: $P b^{2 +} + 2 e^{-} \to P b$
Net reaction: $S n + P b^{2 +} \to P b + S n^{2 +}$
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
$E_{c e l l}^{\circ} = 0.01 V$
$E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.06}{2} log Q$
So,
$0 = 0.01 - \frac{0.06}{2} l o g \frac{[S n^{2 +}]}{[P b^{2 +}]}$
$0.01 = \frac{0.06}{2} l o g \frac{[S n^{2 +}]}{[P b^{2 +}]}$
$l o g \frac{[S n^{2 +}]}{[P b^{2 +}]} = \frac{1}{3}$
$\frac{[S n^{2 +}]}{[P b^{2 +}]} = (10)^{\frac{1}{3}} = 2.154$

Suggest Corrections

Similar questions

Q. For the cell reaction:

P b + S n^{2 +} \to P b^{2 +} + S n

Given that:

P b \to P b^{2 +}, E^{o} = 0.13 V

S n^{2 +} + 2 e^{-} \to S n; E^{o} = - 0.14 V

What would be the ratio of cation concentration for which E = 0?

Q. Calculate

E^{0}

and

E

for the cell

S n | {S n}^{2 +} (1 M) ∥ {P b}^{2 +} (10^{- 3} M) | P b, E^{0} ({S n}^{2 +} / S n) = - 0.14 V, E^{0} ({P b}^{2 +} / P b) = - 0.13 V

Is cell representation is correct?

Q. For the cell reaction

P b + S n^{2 +} \to P b^{2 +} + S n

Calculate the ratio of cation concentration for which E = 0. Given an example where

E^{o}

is zero.
Given that

P b \to P b^{2 +}

;

E^{o}

= 0.13V

S n^{2 +} + 2 e^{-} \to S n

;

E^{o}

= -0.14V

Q. What ratio of

P b^{2 +}

S n^{2 +}

concentration is needed to reverse the following cell reaction?

S n (s) + P b^{2} (a q .) \to S n^{2 +} (a q .) + P b (s)

E_{S n^{2 +} / S n}^{\circ} = - 0.136 v o l t

and

E_{P b^{2 +} / P b}^{\circ} = - 0.126 v o l t

Q. Using Nernst equation for the cell reaction,

P b + S n^{2 +} \to P b^{2 +} + S n

Calculate the ratio

\frac{[P b^{2 +}]}{[S n^{2 +}]}

for which

E_{c e l l} = 0

.
(Given:

E_{P b / P b^{2 +}}^{\circ} = 0.13 v o l t

and

E_{S n^{2 +} / S n}^{\circ} = - 0.14 v o l t

Join BYJU'S Learning Program

For an electrochemical cell Sn(s)|Sn2+(aq.,1M)||Pb2+(aq.,1M)|Pb(s) the ratio [Sn2+][Pb2+] when this cell attains equilibrium is . (Given: E∘Sn2+Sn=−0.14V;E∘Pb2+Pb=−0.13V,2.303RTF=0.06V).

Anodic half: Sn →Sn2++2e− Cathodic half: Pb2++2e−→Pb Net reaction: Sn+Pb2+→Pb+Sn2+ E∘cell=E∘cathode−E∘anode E∘cell=0.01V Ecell=E∘cell−0.062log Q So, 0=0.01−0.062log[Sn2+][Pb2+] 0.01=0.062log[Sn2+][Pb2+] log[Sn2+][Pb2+]=13 [Sn2+][Pb2+]=(10)13=2.154