The standard reaction gibbs energy for the electrochemical reaction- Cr 2 O 72-2 Fe -14 H - 2 Cr 3-2 F e 3-2 H 2 O is -793 kJ mol -1- The standard cell potential is-A- -1-37 VB- -4-11 VC- -2-74 VD- -2-05 V

The correct option is A + 1.37V Δ G^∘ = nFE^∘_cell Here n = 6, 793 × 10^3 = 6 × 96500 × E^∘_cell E^∘_cell = + 1.3696 ≈1.37V ...

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Standard XII

Chemistry

Enthalpy

The standard ...

Question

The standard reaction gibbs energy for the electrochemical reaction.

$C r_{2} O_{7}^{2 -} + 2 F e + 14 H^{+} \to 2 C r^{3 +} + 2 F e^{3 +} + 2 H_{2} O$ is $- 793 k J m o l^{- 1}$ . The standard cell potential is:

+ 1.37V

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+ 4.11V

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+ 2.74V

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+ 2.05V

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Solution

The correct option is A
+ 1.37V

$Δ G^{\circ} = - n F E_{c e l l}^{\circ}$

Here $n = 6$ ,

$- 793 \times 10^{3} = - 6 \times 96500 \times E_{c e l l}^{\circ}$

$E_{c e l l}^{\circ} = + 1.3696$

$\approx 1.37 V$

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Similar questions

Q. The standard reaction, Gibbs energy for the electrochemical reaction.

C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O i s - 793 k J m o l^{- 1}

The standard cell potential is:

The standard reaction Gibbs energy for the electrochemical reaction,

$C r_{2} O_{7}^{2 -} + 2 F e + 14 H^{+} \to 2 C r^{3 +} + 2 F e^{3 +} + 2 H_{2} O$ , is $- 793 k J m o l^{- 1}$ . The standard cell potential is Volts.

C r_{2} O_{7}^{2 -} (a q, 1 M) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q, 1 M) + 7 H_{2} O

. The standard electrode potential for the following reaction is + 1.33V. what is the potential at

p H = 2.0

The cell in which the following reaction occurs:

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

has

E_{c e l l}^{\circ} = 0.236 V

298 K

Calculate the standard Gibbs energy of the cell reaction. (Given :

1 F = 96, 500 C m o l^{- 1})

The cell in which the following reactions occurs:

has = 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

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The standard reaction gibbs energy for the electrochemical reaction. Cr2O2−7+2Fe+14H+→2Cr3++2Fe3++2H2O is −793 kJ mol−1. The standard cell potential is:

The correct option is A + 1.37V ΔG∘=−nFE∘cell Here n=6, −793×103=−6×96500×E∘cell E∘cell=+1.3696 ≈1.37V

The standard reaction gibbs energy for the electrochemical reaction.

$C r_{2} O_{7}^{2 -} + 2 F e + 14 H^{+} \to 2 C r^{3 +} + 2 F e^{3 +} + 2 H_{2} O$ is $- 793 k J m o l^{- 1}$ . The standard cell potential is:

The correct option is A
+ 1.37V

$Δ G^{\circ} = - n F E_{c e l l}^{\circ}$

Here $n = 6$ ,

$- 793 \times 10^{3} = - 6 \times 96500 \times E_{c e l l}^{\circ}$

$E_{c e l l}^{\circ} = + 1.3696$

$\approx 1.37 V$