1996 Iowa Senate election

1996 Iowa Senate election

← 1994 November 5, 1996 1998 →

25 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party
 
Leader Mary Kramer Leonard Boswell
Party Republican Democratic
Leader's seat 37th 44th
(retired)
Last election 23 27
Seats before 23 27
Seats after 29 21
Seat change Increase6 Decrease6

President of the Senate before election

Leonard Boswell
Democratic

Elected President of the Senate

Mary Kramer
Republican

The 1996 Iowa State Senate elections took place as part of the biennial 1996 United States elections. Iowa voters elected state senators in half of the state senate's districts—the 25 even-numbered state senate districts. State senators serve four-year terms in the Iowa State Senate, with half of the seats up for election each cycle. A statewide map of the 50 state Senate districts in the year 1996 is provided by the Iowa General Assembly here.

The primary election on June 4, 1996 determined which candidates appeared on the November 5, 1996 general election ballot. Primary election results can be obtained here.[1] General election results can be obtained here.[2]

Following the previous election, Democrats had control of the Iowa state Senate with 27 seats to Republicans' 23 seats.

To take control of the chamber from Democrats, the Republicans needed to net 3 Senate seats.

Republicans claimed control of the Iowa State Senate following the 1996 general election by netting 6 seats, resulting in Republicans holding 29 seats and Democrats having 21 seats.

  1. ^ "Primary Election 1996 Canvass Summary" (PDF). Iowa Secretary of State. Retrieved April 10, 2020.
  2. ^ "General Election 1996 Canvass Summary" (PDF). Iowa Secretary of State. Retrieved April 14, 2020.

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