A galvanic cell is powered by the following redox reaction: 2Fe + 3(aq) + H2 (g) + 2OH−(aq) → 2Fe + 2(aq) + 2H2O(l)Answer the following questions about this cell. (a) Write a balanced equation for the half-reaction that takes place at the cathode. (b) Write a balanced equation for the half-reaction that takes place at the anode. (c) Calculate the cell voltage under standard conditions.

0.77V

Explanation:

The balanced anode and cathode half reaction equations are shown in the image attached. Reduction occurs at the cathode while oxidation occurs at the anode. Recall that hydrogen is a reference electrode hence its reduction potential is arbitrarily chosen as 0.00V under standard conditions. The hydroxide ion shows that the metal ion was maintained under alkaline conditions.

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

lol for the pointsss