In a Daniel cell,
At cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
At anode: Zn(s) → Zn²⁺(aq) + 2e⁻
Ecell = E°cell - \(\frac{0.0591}{n}\) log \(\frac{anode}{cathode}\)
Zn| ZnSO4 (0.01 M)
CuSO4 | Cu (1.0 M)
In the equation, anode = Zn2+ . Cathode = Cu2+
Substituting the value:
E1 = E° - \(\frac{0.0591}{n}\)log\(\frac{0.01}{1.0}\)
= E° - \(\frac{0.0591}{2} log 10^{-2}\)
E° + 0.0591 V
Now, for E2 : Zn (1.0 M) Cu (0.01)
E2 = E° - \(\frac{0.0591}{2} \)\(log \frac{1}{0.01}\)
E°- \(\frac{0.0591}{2} log^2\)
E° - 0.0591 V
E1 > E2
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion (A) : 02 is liberated in the non-cyclic photophosphorylation.
Reason (R) : Liberation of oxygen is due to photolysis of water.
In the light of the above statements, choose the correct answer from the options given below
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion (A) : The Cro-Magnon man was the direct ancestor of the living modern man.
Reason (R) : Cro-Magnon man had slightly prognathous face.
In the light of the above statements, choose the correct answer from the options given below
This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.