1889 Iowa Senate election

In the 1889 Iowa State Senate elections Iowa voters elected state senators to serve in the twenty-third Iowa General Assembly. Elections were held in 22 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 1889 elections is provided by the Iowa General Assembly here.

The general election took place on November 5, 1889.

Following the previous election, Republicans had control of the Iowa Senate with 32 seats to Democrats' 16 seats and two Independents.

To claim control of the chamber from Republicans, the Democrats needed to net 10 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1889 general election with the balance of power shifting to Republicans holding 28 seats, Democrats having 20 seats, one Independent, and one member of the Union Labor Party (a net gain of 4 seats for Democrats and 1 seat for the Union Labor Party).

Summary of Results

 * Note: The holdover Senators not up for re-election are not listed on this table.

Source:

Detailed Results

 * NOTE: The Iowa Official Register does not contain detailed vote totals for state senate elections in 1889.