What is the voltage of an electrolytic cell with copper and magnesium electrodes?

A. 2.71 V

B. -2.71 V

C. 2.03 V

D. -2.03 V

A. 2.71 V

Explanation:

The standard reduction potential, E° of the metals are as follows:

Mg²⁺ + 2e⁻ ⇌  Mg; E° = -2.372

Cu²⁺ + 2e⁻ ⇌  Cu; E° = +0.337

Since magnesium has the the lower E°, it will serve as the anode in the electrolytic cell while copper will serve as the cathode.

At the anode; Mg ⇌  Mg²⁺ + 2e⁻, E° = -2.372

At the cathode; Cu²⁺ + 2e⁻ ⇌  Cu, E° = +0.337

EMF of the cell = E° cathode - E° anode = 0.337 - (-2.372) = 2.71 V

Therefore, EMF of the cell is 2.71 V

answer: mercury

explanation:

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