The electrical work done during the reaction at 298 K -0-27 V - P - Cl - 2-1 - atm - l A- 420-74 kJ mol -1B- 110-37 kJ mol -1C- 210-37 kJ mol -1D- 105-185 kJ mol -1

The correct option is C 210.37 kJ mol^ 1 Δ G^∘ = electrical work done = n F E^∘_cell = 2 × 96500 × (1.36 0.27)J = 210.37 kJ ...

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

Byju's Answer

Standard VII

Chemistry

Gibb's Energy and Nernst Equation

The electrica...

Question

The electrical work done during the reaction at 298 K:
$2 H g_{(l)} + C l_{2 (g)} \to H g_{2} C l_{2 (s)}, i s G i v e n t h a t E_{\frac{C l_{2}}{C l^{-}}}^{\circ} = 1.36 V; E_{\frac{H g_{2} C l_{2}}{H g, C l^{-}}}^{\circ} = 0.27 V; P_{C l_{2}} = 1 a t m,$

$210.37 k J m o l^{- 1}$

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

$105.185 k J m o l^{- 1}$

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

$420.74 k J m o l^{- 1}$

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

$110.37 k J m o l^{- 1}$

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

Open in App

Solution

The correct option is A
$210.37 k J m o l^{- 1}$

$- Δ G^{\circ} =$ electrical work done $= n F E_{c e l l}^{\circ} = 2 \times 96500 \times (1.36 - 0.27) J = 210.37 k J$

Suggest Corrections

Similar questions

The electrical work done during the reaction at 298 K:
$2 H g_{(l)} + C l_{2 (g)} \to H g_{2} C l_{2 (s)}, i s G i v e n t h a t E_{\frac{C l_{2}}{C l^{-}}}^{\circ} = 1.36 V; E_{\frac{H g_{2} C l_{2}}{H g, C l^{-}}}^{\circ} = 0.27 V; P_{C l_{2}} = 1 a t m,$

Calculate the useful work done during the reaction in kJ/mol.
$A g (S) + \frac{1}{2} C l_{2} (g) \to A g C l (s)$
Given that $E_{C l_{2} / {C l}^{-}}^{0} = + 1.36 V$
and $E_{A g^{+} | A g C l | C l^{-}}^{0} = + 0.220 V$
$P_{C l_{2}} = 1 atm$ and T = 298 K

Q. For the reaction:

4 N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (l)

298 K

. Given that heat of formation of

N H_{3} (g

) and

H_{2} O (l)

are

- 46 k J m o l^{- 1}

and

- 286 k J m o l^{- 1}

respectively. Calculate heat of reaction at

298 K

Q. Question 11

Electrolysis of water is a decomposition reaction. The mole ratio of hydrogen and oxygen gases liberated during electrolysis of water is

(a) 1:1
(b) 2:1
(c) 4:1
(d) 1:2

p² - 1 =

Join BYJU'S Learning Program

The electrical work done during the reaction at 298 K: 2Hg(l)+Cl2(g)→Hg2Cl2(s),isGiven that E∘Cl2Cl−=1.36V;E∘Hg2Cl2Hg,Cl−=0.27V;PCl2=1 atm,

The correct option is A 210.37 kJ mol−1 −ΔG∘= electrical work done =nF E∘cell=2×96500×(1.36−0.27)J=210.37 kJ

The electrical work done during the reaction at 298 K:
$2 H g_{(l)} + C l_{2 (g)} \to H g_{2} C l_{2 (s)}, i s G i v e n t h a t E_{\frac{C l_{2}}{C l^{-}}}^{\circ} = 1.36 V; E_{\frac{H g_{2} C l_{2}}{H g, C l^{-}}}^{\circ} = 0.27 V; P_{C l_{2}} = 1 a t m,$

The correct option is A
$210.37 k J m o l^{- 1}$

$- Δ G^{\circ} =$ electrical work done $= n F E_{c e l l}^{\circ} = 2 \times 96500 \times (1.36 - 0.27) J = 210.37 k J$