The equilibrium constant for the following reaction is 1030- Calculate E- in volts for the cell at 298 K-2X2s - 3Y2-aq-2X23-aq -3Ys

We 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 RTlnK_eq= nFE_cell nbsp; nbsp; ...

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Standard XIII

Chemistry

Equilibrium Constant from Nernst Equation

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

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+0.295 V

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+0.0985 V

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-0.295 V

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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
$- R T l n K_{e q} = - n F E_{c e l l}$
$- R T l n 10^{30} = - n F E_{c e l l}$
$E_{c e l l} = 0.295 V$

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Similar questions

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. The equilibrium constant for the following general reaction is

10^{30}

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E^{0}

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

Q. Calculate the equilibrium constant and free energy change of given following reaction for Daniell cell at 298 K temperature.

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

Cell potential

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(

F = 96500

coulomb)

Q. The equilibrium constant for the reaction given below at 298 K is if

E_{c e l l} = 0.2905 V

at 298 K is:

Z n (s) + F e^{2 +} (a q) 0.01 M \to Z n^{2 +} 0.1 M (a q) + F e (s)

Q. The equilibrium constant for the reaction:

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is 140 at 298 K.
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is:

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