The equation below represents the overall cell reaction for an electrochemical cell that undergoes a spontaneous reaction. fe(s) + pb2+(aq) mc011-1.jpg fe2+(aq) + pb(s) emc011-2.jpgcell = –0.31 v which reaction is the spontaneous reduction reaction? fe(s) mc011-3.jpg fe2+(aq) + 2e– emc011-4.jpg = –0.44 v fe(s) mc011-5.jpg fe2+(aq) + 2e– emc011-6.jpg = +0.44 v pb2+(aq) + 2e– mc011-7.jpg pb(s) emc011-8.jpg = –0.13 v pb2+(aq) + 2e– mc011-9.jpg pb(s) emc011-10.jpg = +0.13 v

Pb2+(aq) + 2e– >Pb(s) = –0.13 V

Explanation:

Given the question asked, we have to examine the reduction potentials of Fe2+ and Pb2+.

Fe2+= -0.44V

Pb2+= -0.13V

Recall that the moire negative the reduction potential of a species, the less likely it is to participate in spontaneous reduction reactions.

From the data before us, Pb2+ will undergo a spontaneous reduction reaction as shown in the answer.

astronomers such as edwin hubble.

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