What is the standard cell voltage for
Cd(s) + 2 Ag+(aq) > Cd2+(aq) + 2 Ag(s) ?
+-o
Ag (aq)+e ---> Ag(s) E =+0.799V
2+ - o
Cd (aq) +2e --->Cd(s) E =-0.402V

a) +0.397 V
b) +0.799 V
c) +1.000 V
d) +1.201 V
e) +2.000 V

Answer : The correct option is, (d) +1.201 V

Explanation :

Here cadmium (Cd) undergoes oxidation by loss of electrons, thus act as anode because less value of electrode potential. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode because more value of electrode potential.

The half-cell reactions are:

Oxidation reaction (anode) :

Reduction reaction (cathode) :

The balanced cell reaction will be,

Given :

Now put all the values in this expression, we get:

Therefore, the standard cell voltage is, +1.201 V

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol