Electrochemistry Part 2
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Question : 16 of 117
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1 Faraday electricity was passed through Cu 2 + ( 1.5 M , 1 L ) ∕ Cu and 0.1 Faraday was passed through Ag + ( 0.2 M , 1 L ) ∕ Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is _ _ _ _ mV (nearest integer) Given : E Cu 2 + ∕ Cu ∘ = 0.34 V
E Ag + ∕ Ag ∘ = 0.8 V
= 0.06 V
[7 Apr 2025 Shift 1]
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