Which equation represents a redox reaction? 1.  2agno3(aq) + cu(s) → cu(no3)2(aq) + 2ag(s) 2.  bacl2(aq) + na2cro4(aq) → bacro4(s) + 2nacl(aq) 3.  h3po4(aq) + 3koh(aq) → 3h2o(l) + k3po4(aq) 4.  caco3(s) → cao(s) + co2(g)?

2agno3(aq) + cu(s) → cu(no3)2(aq) + 2ag(s)

Explanation:

Now let us take a good look at the reaction presented in the answer.

If we look at silver on the lefthand side of the reaction equation, we will realize that silver is in the +2 oxidation state but on the right hand side of the reaction equation, silver is in the zero oxidation state. This implies that silver has been reduced. Reduction refers to a decrease in oxidation number.

Now let us move to copper. Copper is in the zero oxidation state on the lefthand side of the reaction equation but on the right hand side of the reaction equation, it is in the +2 oxidation state. Hence, there is an increase in the oxidation number of copper- oxidation of cooler has occurred. Oxidation rs to a positive increase in oxidation number.

Hence the reaction involves oxidation and reduction, thus it s a redox reaction.

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

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