***50 points*** show all work. 1. what are the 2 methods of balancing redox reactions? 2. practice 8: which atom has a change in oxidation number of +4 in this redox reaction? k2cr2o7 (aq) + h2o (l) + s (s) → koh (aq) + cr2o3 (s) + so2 (g) 3. practice 9: which atom has a change in oxidation number of -3 in this redox reaction? k2cr2o7 (aq) + h2o (l) + s (s) → koh (aq) + cr2o3 (s) + so2 (g) 4. practice 10: balance this redox reaction using the oxidation-number change method and describe each step you used: hno2 (aq) + hi (aq) → no (g) + i2 (s) + h2o (l) 5. practice 11: balance this redox reaction using the oxidation-number change method and describe each step you used: hcl (aq) + zn (s) → zncl2 (s) + h2 (g) 6. do you know how to identify oxidation half-reaction and reduction half-reaction in a redox equation?

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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