1899 Iowa Senate election

1899 Iowa Senate election

← 1897 November 7, 1899 1901 →

31 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party
 
Party Republican Democratic
Last election 39 11
Seats after 35 15
Seat change Decrease4 Increase4

In the 1899 Iowa State Senate elections Iowa voters elected state senators in 31 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 1899 elections is provided by the Iowa General Assembly here.

The 1899 elections occurred before primary elections were established in Iowa by the Primary Election Law in 1907.[1] The general election took place on November 7, 1899.[2]

Following the previous election, Republicans had control of the Iowa Senate with 39 seats to Democrats' 11 seats.

To claim control of the chamber from Republicans, the Democrats needed to net 15 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1899 general election with the balance of power shifting to Republicans holding 35 seats and Democrats having 15 seats (a net gain of 4 seats for Democrats).

  1. ^ "Primary Election Law" (PDF). Iowa General Assembly. Retrieved June 19, 2021.
  2. ^ "General Election 1899 For State Senator" (PDF). Iowa Official Register. Retrieved June 25, 2021.