The equilibrium constant for the following reaction is 1030 - Calculate E - in volts for the cell at 298 K- 2 X 2 s -3 Y 2- aq - 2 X 23- aq -3 Y s A- -0-105 VB- -0-295 VC- -0-0985 VD- -0-295 V

The correct option is B +0.295 VWe know that: Δ G= RTlnK_eq Δ G= nFE_cell At cathode 3Y^2++6e^ → 3Y At anode: 2X_2→ 2X_2^3++6e^ So nbsp;n=6 for reaction RT ...

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

Byju's Answer

Standard XII

Chemistry

Types of Redox Reactions

The equilibri...

Question

The equilibrium constant for the following reaction is $10^{30}$ . Calculate $E^{\circ}$ (in volts) for the cell at 298 K.
$2 X_{2} (s) + 3 Y^{2 +} (a q)$ → $2 X_{2}^{3 +} (a q) + 3 Y (s)$

+0.105 V

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

+0.295 V

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

+0.0985 V

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

-0.295 V

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

Open in App

Solution

The correct option is B +0.295 V
We know that:
$Δ G = - R T l n K_{e q}$
$Δ G = - n F E_{c e l l}$
At cathode
$3 Y^{2 +} + 6 e^{-} \to 3 Y$
At anode:
$2 X_{2} \to 2 X_{2}^{3 +} + 6 e^{-}$
So $n = 6$ for reaction
$- R T l n K_{e q} = - n F E_{c e l l}$
$- R T l n 10^{30} = - n F E_{c e l l}$
$E_{c e l l} = 0.295 V$

Suggest Corrections

Similar questions

Q. The equilibrium constant for the following general reaction is

10^{30}

. Calculate

E^{0}

for the cell at 298K.

2 X_{2} (s) + {3 Y}^{2 +} (a q) ⟶ {2 X}_{2}^{3 +} (a q) + 3 Y (s)

Q. The equilibrium constant for the following reaction is

10^{30}

. Calculate

E^{\circ}

(in volts) for the cell at 298 K.

2 X_{2} (s) + 3 Y^{2 +} (a q)

→

2 X_{2}^{3 +} (a q) + 3 Y (s)

Q. Calculate the equilibrium constant for the reaction,

Z n (s) + A g_{2} O (s) + H_{2} O (l) \to 2 A g (s) + Z n^{2 +} (a q .) + 2 O H^{-} (a q .)

when

E_{c e l l}^{\circ} = 1.11

298 K

Q. The cell in which the following reaction occurs:

2 F e_{(a q)}^{3 +} + 2 I_{(a q)}^{-} \to 2 F e_{(a q)}^{2 +} + I_{2 (s)}

has

E_{c e l l}^{o} = 0.236 V

298 K

.
The equilibrium constant of the cell reaction is:

Q. For the cell reaction,

M g | M g^{2 +} (a q .) ∥ A g^{+} (a q .) | A g

calculate the equilibrium constant at

25^{\circ} C

and maximum work that can be obtained by operating the cell.

E_{M g^{2 +} / M g}^{\circ} = - 2.37 v o l t

and

E_{A g^{+} / A g}^{\circ} = + 0.80 v o l t

Join BYJU'S Learning Program

The equilibrium constant for the following reaction is 1030. Calculate E∘ (in volts) for the cell at 298 K. 2X2(s)+3Y2+(aq)→2X3+2(aq)+3Y(s)

The correct option is B +0.295 VWe know that: ΔG=−RTlnKeq ΔG=−nFEcell At cathode 3Y2++6e−→3Y At anode: 2X2→2X3+2+6e− So n=6 for reaction −RTlnKeq=−nFEcell −RTln1030=−nFEcell Ecell=0.295V

The equilibrium constant for the following reaction is $10^{30}$ . Calculate $E^{\circ}$ (in volts) for the cell at 298 K.
$2 X_{2} (s) + 3 Y^{2 +} (a q)$ → $2 X_{2}^{3 +} (a q) + 3 Y (s)$