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Nernst Equation

A galvanic ce...

Question

A galvanic cell consists of a strip of cobalt metal dipping into 1.0 M $C o^{2 +}$ ions, and another half cell in which a piece of platinum dips into a 1.0 M solution of $C l^{-}$ ions. Now $C l_{2} (g)$ at a pressure of 1.0 atm is passed into this solution. The observed cell voltage is 1.63 V, and as the cell reaction proceeds, the Co electrode becomes negative.
$E^{⊖} (1 / 2 C l_{2} + e^{-} \to C l^{⊖}) = 1.36 V$
The cell voltage (in V) if the $[C o^{2 +}]$ were reduced to 0.01 M will be (multiply answer by 10 and write nearest integer value) :

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Solution

$E_{c e l l} = 1.63 - \frac{0.059}{2} l o g$ $(0.01) = 1.63 + 0.059 = 1.689$ V.
Answer $= 10 \times 1.689 = 16.89 \approx 17$

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A galvanic cell consists of a strip of cobalt metal dipping into 1.0 M

C o^{2 +}

ions, and another half cell in which a piece of platinum dips into a 1.0 M solution of

C l^{-}

C l_{2} (g)

at a pressure of 1.0 atm is passed into this solution. The observed cell voltage is 1.63 V, and as the cell reaction proceeds, the Co electrode becomes negative.

E^{⊖} (1 / 2 C l_{2} + e^{-} \to C l^{⊖}) = 1.36 V

Assertion: The cell voltage increases, if pressure of

C l_{2} (g)

Reason: Difference in cell voltage is directly proportional to the pressure of

C l_{2} (g)

.

Q. A galvanic cell consists of a strip of cobalt metal dipping into 1.0 M

C o^{2 +}

ions, and another half cell in which a piece of platinum dips into a 1.0 M solution of

C l^{-}

C l_{2} (g)

at a pressure of 1.0 atm is passed into this solution. The observed cell voltage is 1.63 V, and as the cell reaction proceeds, the Co electrode becomes negative.

E^{⊖} (1 / 2 C l_{2} + e^{-} \to C l^{⊖}) = 1.36 V

E^{⊖}

of cobalt electrode is:

Q. Consider the following electrode processes of a cell,

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