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निम्नलिखित सेल की दक्षता 84% है, तो सेल ...

निम्नलिखित सेल की दक्षता 84% है, तो सेल का मानक इलेक्ट्रोड विभव कितना होगा?
`A(s) + B^(2+) (aq) = A^(2+) (aq) + B(s), DeltaH = -285 kJ`

A

1.20 V

B

2.40 V

C

1.10 V

D

1.24 V

Text Solution

Verified by Experts

The correct Answer is:
D

दक्षता `=(DeltaG^(@))/(DeltaH^(@)) =(-nFE^(@))/(DeltaH^(+))`
`0.84 = (-2 xx E^(@) xx 96500)/(-285 xx 1000)`
`E^(@) = +1.24 V`
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The efficiency of a hypothetical cell is about 84% which involves the following reactions: A(s) + B^(2+)(aq) rightarrow A^(2+)(aq) _ B(s) DeltaH = -285kJ Then, the standard electrode potential of the cell will be: (Asume DeltaS = 0)