Find the potentials of the following electrochemical cell:
cd | cd2+, m = 0.10 ‖ ni2+, m = 0.50 | ni

0.18 V

Explanation:-

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

=-0.40V[/tex]

=-0.24V[/tex]

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

Where both are standard reduction potentials.

Using Nernst equation :

where,

n = number of electrons in oxidation-reduction reaction = 2

= standard electrode potential = 0.16 V

Thus the potential of the following electrochemical cell is 0.18 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