Given the equilibrium constant, k c of the reaction : c u ( s ) 2 a g ( a q ) → c u 2 ( a q ) 2 a g ( s ) is 10 × 10 15 , calculate the e o c e l l of this reaction at 298 k [ 2.303 r t f a t 298 k = 0.059 v ]. Write the equilibrium constant (kc) expression for the following reactions. (i) cu^2 (aq) 2ag(s) ⇌ cu(s) 2ag^ (aq) asked oct 12, 2020 in physical and chemical equilibrium by manish01 ( 45.7k points).
What will be the value of log k e q for the given reaction at 298 k? where, k e q is the equilibrium constant for the given reaction c u (s) 2 a g (a q) → c u 2 (a q) 2 a g (s); e 0 c e l l = 0.46 v. Ncert example page no. 74 electrochemistryproblem 3.2: calculate the equilibrium constant of the reaction:cu (s) 2a. Given the equilibrium constant (k c) of the reaction: cu(s) 2ag (aq) → cu 2 (aq) 2ag(s) is 10 × 10 15, calculate the \(e{^\circ {cell}}\) of this reaction at 298 k. \(\left[ {2.303\frac{{rt}}{f}at\;298\;k = 0.059\;v} \right]\) 1. 0.4736 v 2. 0.4736 mv 3. 0.4736 mv 4. 0.04736 v. The equilibrium constant (k) for the above reaction will be:. [ given, e ∘ c e l l = 0.46 v] view solution.
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