1891 Iowa Senate election

1891 Iowa Senate election

← 1889 November 3, 1891 1893 →

32 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party Third party
 
Party Democratic Republican Populist
Last election 20 28 0
Seats after 25 24 1[a]
Seat change Increase5 Decrease4 Increase1

  Fourth party Fifth party
 
Party Union Labor Independent
Last election 1 1
Seats after 0[a] 0
Seat change Decrease1 Decrease1

In the 1891 Iowa State Senate elections Iowa voters elected state senators to serve in the twenty-fourth Iowa General Assembly. Elections were held in 32 of the state senate's 50 districts. State senators serve four-year terms in the Iowa State Senate.

A statewide map of the 50 state Senate districts in the 1891 elections is provided by the Iowa General Assembly here.

The general election took place on November 3, 1891.[1]

Following the previous election, Republicans had control of the Iowa Senate with 28 seats to Democrats' 20 seats, one seat for the Union Labor Party[a] and one Independent.

To claim control of the chamber from Republicans, the Democrats needed to net 5 Senate seats.

Democrats claimed control of the Iowa State Senate following the 1891 general election with the balance of power shifting to Democrats holding 25 seats, Republicans having 24 seats, and a lone seat for the People's Party[a] (a net gain of 5 seats for Democrats and 1 seat for the People's Party).


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  1. ^ "The Twenty-Fourth General Assembly. SENATE" (PDF). Iowa Official Register. Retrieved June 26, 2021.

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